Question

In Fig. 13.30, ABC is an equilateral triangle
and AD is perpendicular to BC. Prove that
∆ADB = ∆ADC in three different
ways.​
please answer it fast i wil mark as brainliest

Answer 1

Step-by-step explanation:

In △ ABC,

AB = BC = CA,

AD ⊥ BC, BE ⊥ AC,

Proof: In △ ADC and △ BEC

∠ADC = ∠BEC (each 90°)

∠ACD = ∠BCE (common)

and AC = BC (Sides of an equilateral triangle)

∴ △ ADC ≅ △ BEC (A.A.S. Axiom)

Hence, (i) AD = BE (c.p.c.t)

And (ii) BD = CE (c.p.c.t)

Hence proved.

Answer 2

According To Question:-

In Fig. 13.30, ABC is an equilateral triangle

and AD is perpendicular to BC. Prove that

∆ADB = ∆ADC in three different

ways.

Solution:-

In △ADB and △ADC

AB=AC (given)

AD=AD (common side)

∠ADB = ∠ADC = 90°

SSA congruency

Hence,△ADB≅△ADC by RHS property