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edited 17 hours ago

Ankit Kumar

We place the triangle on coordinate axes for convenience.

Let $A = (0,0)$, $B = (2b,0)$, $C = (0,2c)$, so that $\angle A = 90^\circ$.

Step 1: Equation of the median $BD$

The midpoint of $AC$ is: $$ \left(0, c\right) $$

So the median from $B$ passes through $(2b,0)$ and $(0,c)$.

Using intercept form: $$ \frac{x}{2b} + \frac{y}{c} = 1 $$

Multiplying throughout by $2bc$: $$ cx + 2by = 2bc \quad \text{...(i)} $$

Step 2: Coordinates of $D$ using perpendicular condition

Given $AD \perp BD$, point $D$ is the foot of the perpendicular from $A(0,0)$ to the line (i).

For a line $ax + by + d = 0$, the foot of perpendicular from $(x_0,y_0)$$(0,0)$ is: $$ \left( \frac{-ad}{a^2 + b^2},; \frac{-bd}{a^2 + b^2} \right) $$

Rewrite (i): $$ cx + 2by - 2bc = 0 $$

So:

$a = c$
$b = 2b$
$d = -2bc$

$(x0,y0)=(0,0)=vertex A$

Substituting: $$ x = \frac{-c(-2bc)}{c^2 + (2b)^2} = \frac{2bc^2}{c^2 + 4b^2} $$

$$ y = \frac{-(2b)(-2bc)}{c^2 + 4b^2} = \frac{4b^2 c}{c^2 + 4b^2} $$

So, $$ D = \left( \frac{2bc^2}{c^2 + 4b^2},; \frac{4b^2 c}{c^2 + 4b^2} \right) \quad \text{...(ii)} $$

Step 3: Equation of angle bisector of $\angle C$

Using the internal angle bisector theorem at $C$:

Points on the angle bisector divide opposite side in the ratio of adjacent sides: $$ \frac{AF}{FB} = \frac{AC}{CB} $$

Now: $$ AC = 2c,\quad CB = \sqrt{(2b)^2 + (2c)^2} = 2\sqrt{b^2 + c^2} $$

So: $$ \frac{AF}{FB} = \frac{2c}{2\sqrt{b^2 + c^2}} = \frac{c}{\sqrt{b^2 + c^2}} $$

Let $F = (x,0)$ on $AB$.

Then: $$ \frac{x}{2b - x} = \frac{c}{\sqrt{b^2 + c^2}} $$

Cross-multiplying: $$ x\sqrt{b^2 + c^2} = c(2b - x) $$

$$ x\sqrt{b^2 + c^2} = 2bc - cx $$

$$ x(\sqrt{b^2 + c^2} + c) = 2bc $$

$$ x = \frac{2bc}{\sqrt{b^2 + c^2} + c} $$

So the x-intercept of the angle bisector is: $$ AF = \frac{2bc}{\sqrt{b^2 + c^2} + c} $$

Thus, equation in intercept form: $$ \frac{x}{\frac{2bc}{\sqrt{b^2 + c^2} + c}} + \frac{y}{2c} = 1 $$

Rewriting: $$ \frac{x(\sqrt{b^2 + c^2} + c)}{2bc} + \frac{y}{2c} = 1 \quad \text{...(iii)} $$

Step 4: Substitute $D$ into (iii)

Substitute coordinates from (ii):

First term: $$ \frac{x(\sqrt{b^2 + c^2} + c)}{2bc} = \frac{\frac{2bc^2}{c^2 + 4b^2}(\sqrt{b^2 + c^2} + c)}{2bc} $$

Cancel $2bc$: $$ = \frac{c(\sqrt{b^2 + c^2} + c)}{c^2 + 4b^2} $$

Second term: $$ \frac{y}{2c} = \frac{\frac{4b^2 c}{c^2 + 4b^2}}{2c} = \frac{2b^2}{c^2 + 4b^2} $$

So equation becomes: $$ \frac{c(\sqrt{b^2 + c^2} + c)}{c^2 + 4b^2} + \frac{2b^2}{c^2 + 4b^2} = 1 $$

Combine: $$ \frac{c(\sqrt{b^2 + c^2} + c) + 2b^2}{c^2 + 4b^2} = 1 $$

Multiply both sides: $$ c(\sqrt{b^2 + c^2} + c) + 2b^2 = c^2 + 4b^2 $$

Expand: $$ c\sqrt{b^2 + c^2} + c^2 + 2b^2 = c^2 + 4b^2 $$

Cancel $c^2$: $$ c\sqrt{b^2 + c^2} + 2b^2 = 4b^2 $$

$$ c\sqrt{b^2 + c^2} = 2b^2 $$

Step 5: Final simplification

Square both sides: $$ c^2(b^2 + c^2) = 4b^4 $$

$$ b^2c^2 + c^4 = 4b^4 $$

Divide by $b^4$, let $\eta = \frac{c}{b}$: $$ \eta^2 + \eta^4 = 4 $$

Rearrange: $$ \eta^4 + \eta^2 - 4 = 0 $$

Step 6: Solve quadratic

Let $u = \eta^2$: $$ u^2 + u - 4 = 0 $$

$$ u = \frac{-1 \pm \sqrt{17}}{2} $$

Since $\eta > 0$: $$ \eta = \sqrt{\frac{-1 + \sqrt{17}}{2}} $$

✅ Final Result

$$ \boxed{ \frac{c}{b} = \sqrt{\frac{\sqrt{17} - 1}{2}} } $$

The rest is easy, coordinates of D,A,B are known, you can check the distances. They are messy, but are equal.

We place the triangle on coordinate axes for convenience.

Let $A = (0,0)$, $B = (2b,0)$, $C = (0,2c)$, so that $\angle A = 90^\circ$.

Step 1: Equation of the median $BD$

The midpoint of $AC$ is: $$ \left(0, c\right) $$

So the median from $B$ passes through $(2b,0)$ and $(0,c)$.

Using intercept form: $$ \frac{x}{2b} + \frac{y}{c} = 1 $$

Multiplying throughout by $2bc$: $$ cx + 2by = 2bc \quad \text{...(i)} $$

Step 2: Coordinates of $D$ using perpendicular condition

Given $AD \perp BD$, point $D$ is the foot of the perpendicular from $A(0,0)$ to the line (i).

For a line $ax + by + d = 0$, the foot of perpendicular from $(x_0,y_0)$ is: $$ \left( \frac{-ad}{a^2 + b^2},; \frac{-bd}{a^2 + b^2} \right) $$

Rewrite (i): $$ cx + 2by - 2bc = 0 $$

So:

$a = c$
$b = 2b$
$d = -2bc$

$(x0,y0)=(0,0)=vertex A$

Substituting: $$ x = \frac{-c(-2bc)}{c^2 + (2b)^2} = \frac{2bc^2}{c^2 + 4b^2} $$

$$ y = \frac{-(2b)(-2bc)}{c^2 + 4b^2} = \frac{4b^2 c}{c^2 + 4b^2} $$

So, $$ D = \left( \frac{2bc^2}{c^2 + 4b^2},; \frac{4b^2 c}{c^2 + 4b^2} \right) \quad \text{...(ii)} $$

Step 3: Equation of angle bisector of $\angle C$

Using the internal angle bisector theorem at $C$:

Points on the angle bisector divide opposite side in the ratio of adjacent sides: $$ \frac{AF}{FB} = \frac{AC}{CB} $$

Now: $$ AC = 2c,\quad CB = \sqrt{(2b)^2 + (2c)^2} = 2\sqrt{b^2 + c^2} $$

So: $$ \frac{AF}{FB} = \frac{2c}{2\sqrt{b^2 + c^2}} = \frac{c}{\sqrt{b^2 + c^2}} $$

Let $F = (x,0)$ on $AB$.

Then: $$ \frac{x}{2b - x} = \frac{c}{\sqrt{b^2 + c^2}} $$

Cross-multiplying: $$ x\sqrt{b^2 + c^2} = c(2b - x) $$

$$ x\sqrt{b^2 + c^2} = 2bc - cx $$

$$ x(\sqrt{b^2 + c^2} + c) = 2bc $$

$$ x = \frac{2bc}{\sqrt{b^2 + c^2} + c} $$

So the x-intercept of the angle bisector is: $$ AF = \frac{2bc}{\sqrt{b^2 + c^2} + c} $$

Thus, equation in intercept form: $$ \frac{x}{\frac{2bc}{\sqrt{b^2 + c^2} + c}} + \frac{y}{2c} = 1 $$

Rewriting: $$ \frac{x(\sqrt{b^2 + c^2} + c)}{2bc} + \frac{y}{2c} = 1 \quad \text{...(iii)} $$

Step 4: Substitute $D$ into (iii)

Substitute coordinates from (ii):

First term: $$ \frac{x(\sqrt{b^2 + c^2} + c)}{2bc} = \frac{\frac{2bc^2}{c^2 + 4b^2}(\sqrt{b^2 + c^2} + c)}{2bc} $$

Cancel $2bc$: $$ = \frac{c(\sqrt{b^2 + c^2} + c)}{c^2 + 4b^2} $$

Second term: $$ \frac{y}{2c} = \frac{\frac{4b^2 c}{c^2 + 4b^2}}{2c} = \frac{2b^2}{c^2 + 4b^2} $$

So equation becomes: $$ \frac{c(\sqrt{b^2 + c^2} + c)}{c^2 + 4b^2} + \frac{2b^2}{c^2 + 4b^2} = 1 $$

Combine: $$ \frac{c(\sqrt{b^2 + c^2} + c) + 2b^2}{c^2 + 4b^2} = 1 $$

Multiply both sides: $$ c(\sqrt{b^2 + c^2} + c) + 2b^2 = c^2 + 4b^2 $$

Expand: $$ c\sqrt{b^2 + c^2} + c^2 + 2b^2 = c^2 + 4b^2 $$

Cancel $c^2$: $$ c\sqrt{b^2 + c^2} + 2b^2 = 4b^2 $$

$$ c\sqrt{b^2 + c^2} = 2b^2 $$

Step 5: Final simplification

Square both sides: $$ c^2(b^2 + c^2) = 4b^4 $$

$$ b^2c^2 + c^4 = 4b^4 $$

Divide by $b^4$, let $\eta = \frac{c}{b}$: $$ \eta^2 + \eta^4 = 4 $$

Rearrange: $$ \eta^4 + \eta^2 - 4 = 0 $$

Step 6: Solve quadratic

Let $u = \eta^2$: $$ u^2 + u - 4 = 0 $$

$$ u = \frac{-1 \pm \sqrt{17}}{2} $$

Since $\eta > 0$: $$ \eta = \sqrt{\frac{-1 + \sqrt{17}}{2}} $$

✅ Final Result

$$ \boxed{ \frac{c}{b} = \sqrt{\frac{\sqrt{17} - 1}{2}} } $$

The rest is easy, coordinates of D,A,B are known, you can check the distances. They are messy, but are equal.

We place the triangle on coordinate axes for convenience.

Let $A = (0,0)$, $B = (2b,0)$, $C = (0,2c)$, so that $\angle A = 90^\circ$.

Step 1: Equation of the median $BD$

The midpoint of $AC$ is: $$ \left(0, c\right) $$

So the median from $B$ passes through $(2b,0)$ and $(0,c)$.

Using intercept form: $$ \frac{x}{2b} + \frac{y}{c} = 1 $$

Multiplying throughout by $2bc$: $$ cx + 2by = 2bc \quad \text{...(i)} $$

Step 2: Coordinates of $D$ using perpendicular condition

Given $AD \perp BD$, point $D$ is the foot of the perpendicular from $A(0,0)$ to the line (i).

For a line $ax + by + d = 0$, the foot of perpendicular from $(0,0)$ is: $$ \left( \frac{-ad}{a^2 + b^2},; \frac{-bd}{a^2 + b^2} \right) $$

Rewrite (i): $$ cx + 2by - 2bc = 0 $$

So:

$a = c$
$b = 2b$
$d = -2bc$

Substituting: $$ x = \frac{-c(-2bc)}{c^2 + (2b)^2} = \frac{2bc^2}{c^2 + 4b^2} $$

$$ y = \frac{-(2b)(-2bc)}{c^2 + 4b^2} = \frac{4b^2 c}{c^2 + 4b^2} $$

So, $$ D = \left( \frac{2bc^2}{c^2 + 4b^2},; \frac{4b^2 c}{c^2 + 4b^2} \right) \quad \text{...(ii)} $$

Step 3: Equation of angle bisector of $\angle C$

Using the internal angle bisector theorem at $C$:

Points on the angle bisector divide opposite side in the ratio of adjacent sides: $$ \frac{AF}{FB} = \frac{AC}{CB} $$

Now: $$ AC = 2c,\quad CB = \sqrt{(2b)^2 + (2c)^2} = 2\sqrt{b^2 + c^2} $$

So: $$ \frac{AF}{FB} = \frac{2c}{2\sqrt{b^2 + c^2}} = \frac{c}{\sqrt{b^2 + c^2}} $$

Let $F = (x,0)$ on $AB$.

Then: $$ \frac{x}{2b - x} = \frac{c}{\sqrt{b^2 + c^2}} $$

Cross-multiplying: $$ x\sqrt{b^2 + c^2} = c(2b - x) $$

$$ x\sqrt{b^2 + c^2} = 2bc - cx $$

$$ x(\sqrt{b^2 + c^2} + c) = 2bc $$

$$ x = \frac{2bc}{\sqrt{b^2 + c^2} + c} $$

So the x-intercept of the angle bisector is: $$ AF = \frac{2bc}{\sqrt{b^2 + c^2} + c} $$

Thus, equation in intercept form: $$ \frac{x}{\frac{2bc}{\sqrt{b^2 + c^2} + c}} + \frac{y}{2c} = 1 $$

Rewriting: $$ \frac{x(\sqrt{b^2 + c^2} + c)}{2bc} + \frac{y}{2c} = 1 \quad \text{...(iii)} $$

Step 4: Substitute $D$ into (iii)

Substitute coordinates from (ii):

First term: $$ \frac{x(\sqrt{b^2 + c^2} + c)}{2bc} = \frac{\frac{2bc^2}{c^2 + 4b^2}(\sqrt{b^2 + c^2} + c)}{2bc} $$

Cancel $2bc$: $$ = \frac{c(\sqrt{b^2 + c^2} + c)}{c^2 + 4b^2} $$

Second term: $$ \frac{y}{2c} = \frac{\frac{4b^2 c}{c^2 + 4b^2}}{2c} = \frac{2b^2}{c^2 + 4b^2} $$

So equation becomes: $$ \frac{c(\sqrt{b^2 + c^2} + c)}{c^2 + 4b^2} + \frac{2b^2}{c^2 + 4b^2} = 1 $$

Combine: $$ \frac{c(\sqrt{b^2 + c^2} + c) + 2b^2}{c^2 + 4b^2} = 1 $$

Multiply both sides: $$ c(\sqrt{b^2 + c^2} + c) + 2b^2 = c^2 + 4b^2 $$

Expand: $$ c\sqrt{b^2 + c^2} + c^2 + 2b^2 = c^2 + 4b^2 $$

Cancel $c^2$: $$ c\sqrt{b^2 + c^2} + 2b^2 = 4b^2 $$

$$ c\sqrt{b^2 + c^2} = 2b^2 $$

Step 5: Final simplification

Square both sides: $$ c^2(b^2 + c^2) = 4b^4 $$

$$ b^2c^2 + c^4 = 4b^4 $$

Divide by $b^4$, let $\eta = \frac{c}{b}$: $$ \eta^2 + \eta^4 = 4 $$

Rearrange: $$ \eta^4 + \eta^2 - 4 = 0 $$

Step 6: Solve quadratic

Let $u = \eta^2$: $$ u^2 + u - 4 = 0 $$

$$ u = \frac{-1 \pm \sqrt{17}}{2} $$

Since $\eta > 0$: $$ \eta = \sqrt{\frac{-1 + \sqrt{17}}{2}} $$

✅ Final Result

$$ \boxed{ \frac{c}{b} = \sqrt{\frac{\sqrt{17} - 1}{2}} } $$

The rest is easy, coordinates of D,A,B are known, you can check the distances. They are messy, but are equal.

Source Link

answered 17 hours ago

Ankit Kumar

We place the triangle on coordinate axes for convenience.

Let $A = (0,0)$, $B = (2b,0)$, $C = (0,2c)$, so that $\angle A = 90^\circ$.

Step 1: Equation of the median $BD$

The midpoint of $AC$ is: $$ \left(0, c\right) $$

So the median from $B$ passes through $(2b,0)$ and $(0,c)$.

Using intercept form: $$ \frac{x}{2b} + \frac{y}{c} = 1 $$

Multiplying throughout by $2bc$: $$ cx + 2by = 2bc \quad \text{...(i)} $$

Step 2: Coordinates of $D$ using perpendicular condition

Given $AD \perp BD$, point $D$ is the foot of the perpendicular from $A(0,0)$ to the line (i).

For a line $ax + by + d = 0$, the foot of perpendicular from $(x_0,y_0)$ is: $$ \left( \frac{-ad}{a^2 + b^2},; \frac{-bd}{a^2 + b^2} \right) $$

Rewrite (i): $$ cx + 2by - 2bc = 0 $$

So:

$a = c$
$b = 2b$
$d = -2bc$
$(x0,y0)=(0,0)=vertex A$

Substituting: $$ x = \frac{-c(-2bc)}{c^2 + (2b)^2} = \frac{2bc^2}{c^2 + 4b^2} $$

$$ y = \frac{-(2b)(-2bc)}{c^2 + 4b^2} = \frac{4b^2 c}{c^2 + 4b^2} $$

So, $$ D = \left( \frac{2bc^2}{c^2 + 4b^2},; \frac{4b^2 c}{c^2 + 4b^2} \right) \quad \text{...(ii)} $$

Step 3: Equation of angle bisector of $\angle C$

Using the internal angle bisector theorem at $C$:

Points on the angle bisector divide opposite side in the ratio of adjacent sides: $$ \frac{AF}{FB} = \frac{AC}{CB} $$

Now: $$ AC = 2c,\quad CB = \sqrt{(2b)^2 + (2c)^2} = 2\sqrt{b^2 + c^2} $$

So: $$ \frac{AF}{FB} = \frac{2c}{2\sqrt{b^2 + c^2}} = \frac{c}{\sqrt{b^2 + c^2}} $$

Let $F = (x,0)$ on $AB$.

Then: $$ \frac{x}{2b - x} = \frac{c}{\sqrt{b^2 + c^2}} $$

Cross-multiplying: $$ x\sqrt{b^2 + c^2} = c(2b - x) $$

$$ x\sqrt{b^2 + c^2} = 2bc - cx $$

$$ x(\sqrt{b^2 + c^2} + c) = 2bc $$

$$ x = \frac{2bc}{\sqrt{b^2 + c^2} + c} $$

So the x-intercept of the angle bisector is: $$ AF = \frac{2bc}{\sqrt{b^2 + c^2} + c} $$

Thus, equation in intercept form: $$ \frac{x}{\frac{2bc}{\sqrt{b^2 + c^2} + c}} + \frac{y}{2c} = 1 $$

Rewriting: $$ \frac{x(\sqrt{b^2 + c^2} + c)}{2bc} + \frac{y}{2c} = 1 \quad \text{...(iii)} $$

Step 4: Substitute $D$ into (iii)

Substitute coordinates from (ii):

First term: $$ \frac{x(\sqrt{b^2 + c^2} + c)}{2bc} = \frac{\frac{2bc^2}{c^2 + 4b^2}(\sqrt{b^2 + c^2} + c)}{2bc} $$

Cancel $2bc$: $$ = \frac{c(\sqrt{b^2 + c^2} + c)}{c^2 + 4b^2} $$

Second term: $$ \frac{y}{2c} = \frac{\frac{4b^2 c}{c^2 + 4b^2}}{2c} = \frac{2b^2}{c^2 + 4b^2} $$

So equation becomes: $$ \frac{c(\sqrt{b^2 + c^2} + c)}{c^2 + 4b^2} + \frac{2b^2}{c^2 + 4b^2} = 1 $$

Combine: $$ \frac{c(\sqrt{b^2 + c^2} + c) + 2b^2}{c^2 + 4b^2} = 1 $$

Multiply both sides: $$ c(\sqrt{b^2 + c^2} + c) + 2b^2 = c^2 + 4b^2 $$

Expand: $$ c\sqrt{b^2 + c^2} + c^2 + 2b^2 = c^2 + 4b^2 $$

Cancel $c^2$: $$ c\sqrt{b^2 + c^2} + 2b^2 = 4b^2 $$

$$ c\sqrt{b^2 + c^2} = 2b^2 $$

Step 5: Final simplification

Square both sides: $$ c^2(b^2 + c^2) = 4b^4 $$

$$ b^2c^2 + c^4 = 4b^4 $$

Divide by $b^4$, let $\eta = \frac{c}{b}$: $$ \eta^2 + \eta^4 = 4 $$

Rearrange: $$ \eta^4 + \eta^2 - 4 = 0 $$

Step 6: Solve quadratic

Let $u = \eta^2$: $$ u^2 + u - 4 = 0 $$

$$ u = \frac{-1 \pm \sqrt{17}}{2} $$

Since $\eta > 0$: $$ \eta = \sqrt{\frac{-1 + \sqrt{17}}{2}} $$

✅ Final Result

$$ \boxed{ \frac{c}{b} = \sqrt{\frac{\sqrt{17} - 1}{2}} } $$

The rest is easy, coordinates of D,A,B are known, you can check the distances. They are messy, but are equal.