Question

IN FIG., ANGLE ABC = ANGLE ACB. BD IS PERPENDICULAR TO AB AND CD IS PERPENDICULAR TO AC. PROVE THAT BD = CD.​

Answer 1

Step-by-step explanation:

Given:- In ΔABC, ∠ABC = ∠ACB and BD is perpendicular to AB, also CD is perpendicular to AC.

To Prove:- BD = CD

Proof:-

We know that,

∠ABC = ∠ACB ----- 1

BD is perpendicular to AB

So,∠ABD = 90° ----- 2

CD is perpendicular to AC

So, ∠ACD = 90° ----- 3

Then, from eq.2 and eq.3 we get,

∠ABD = ∠ACD = 90° ----- 4

Now,

∠ABD = ∠ABC + ∠CBD ----- 5

also, ∠ACD = ∠ACB + ∠BCD ----- 6

Now, from eq.4 we get,

∠ABD = ∠ACD

also, from eq.5 and eq.6 we get

∠ABC + ∠CBD = ∠ACB + ∠BCD

But, from eq.1 we know that,

∠ABC = ∠ACB

then,

∠ABC + ∠CBD = ∠ABC + ∠BCD (because ∠ABC = ∠ ACB)

∠ABC + ∠CBD - ∠ABC = ∠BCD

∠CBD = ∠BCD

Then,

CD = BD (Properties of an Isosceles Δ)

Hence proved

Hope it helped and you understood it........All the best