Question

ABC is a triangle in which altitudes BD and CE to sides AC and AB are equal show that triangle ABD is congruent triangle ACE and AB=AC i.e., ABC is an isosceles triangle

Answer 1

Shown that Δ ABD ≅ ΔACE , AB = AC , ABC is an isosceles triangle

Step-by-step explanation:

ABC is a triangle in which altitudes BD and CE to sides AC and AB are equal

Method 1 :

Area of Triangle ABC

= (1/2) * AB * CE = (1/2) AC * BD

BD = CE given

=> AB = AC

=> Δ ABC is isosceles

now in Δ ABD & ΔACE

AB = AC

BD = CE

∠ADB = ∠AEC = 90°

=>  Δ ABD ≅ ΔACE

Method 2 :

in Δ BCE & CBD

BC = CB  ( common)

CE = BD

∠BEC = ∠CDB = 90°

=> Δ BCE ≅ CBD

=> BE = CD

=> ∠B = ∠C

=> AC = AB

=> Δ ABC is isosceles

AE = AB - BE

DE = AC - CD

=> AE = DE

now in Δ ABD & ΔACE

AB = AC

BD = CE

AE = DE

=>  Δ ABD ≅ ΔACE

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Answer 2

Hence Proved.

Step-by-step explanation:

Given,

$\triangle ABC$ where altitudes $BD$ and $CE$ to sides $AC$ and $AB$ are equal.

Area of $\triangle ABC$ when $AB$ as base,

$=\frac{1}{2}\times AB\times CE$__1

Also,

Area of $\triangle ABC$ when $AC$ as base,

$=\frac{1}{2}\times AC\times BD$__2

Equation-1&2 are equal,

$\frac{1}{2}\times AB\times CE=\frac{1}{2}\times AC\times BD$

∴$AB=AC$ (∵ $BD=CE$)

∴ $\triangle ABC$ is isosceles triangle.

For $\triangle ABD\ and\ \triangle ACE$,  

• $AB=AC$
• $BD=CE$
• $\angle A$ is common.

∴ $\triangle ABD$≅$\triangle ACE$

Hence Proved.