Question

AD and BE are respectively the altitudes of triangle ABC, in which, AC=BC. Prove AE=BD

Answer 1

Answer:

AD and BE are altitudes of an isosceles triangle ABC with AC = BC. Prove that AE = BD.

ΔCAD and ΔCBE

CA = CB ...(Isosceles triangles)

∠CDA = ∠CEB = 90°

∠ACD = ∠BCE = ...(common)

Therefore, ΔCAD ≅ ΔCBE ...(AAS criteria)

Hence, CE = CD

But, CA = CB

⇒ AE + CE = BD + CD

⇒ AE = BD.

Step-by-step explanation:

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Answer 2

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GiveN:

• AD and BE are altitudes on BC and AC respectively.
• AC = BC

We have to prove that AE = BD!

Let's check in ∆ACD and ∆BCE,

1. CA = CB (Given in the Q.)
2. Angle CDA = Angle CEB (90°)
3. Angle ACD = Angle BCE (common)

⇛ ∆ACD is congurent to ∆BCE (AAS congurence)

Now, CE = CD (By CPCT)

We know, AC = BC (As Provided in the Q.)

⇛ AE + EC = BD + DC

⇛ AE = BD (as CE = CD proved!)

$\therefore$Hence proved!