Question

ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively. Then:​

Answer 1

Answer :

$\large\fbox{ \green\bf Corrected Question :}$

ABC is an isosceles triangle in which altitudes BE & CF are drawn

to sides Ac and AB respectively . Show that these altitudes are  equal

Step-by-step explanation :

$\large\fbox { \orange\bf Given :}$

ΔABC is an isosceles triangle

∴  AB = AC

$\large\fbox { \blue\bf To prove :}$

BE = CF

$\large\fbox { \red\bf Proof :}$

⇒ ∠ACB = ∠ABC  { Angles opposite to equal sides are equal }

In ΔBEC and ΔCFB , we have

• ∠EBC = ∠FCB { Proved }

• BC = CB { Common }

• ∠BEC = ∠CFB { Each 90° }

Therefore , ΔBEC ≌ ΔCFB  → { by using ASA criterian }

BE = CF { CPCT }

#Mbappe007

Mark as brainliest if you liked this answer

Answer 2

Answer:

• BE = CF ( By C P C T )

Step-by-step explanation:

★ GIVEN :

• ΔABC is an isosceles triangle

★ TO PROVE :

• BE = CF

★ SOLUTION :

• We need to about RHS Congruency to answer these.

• In ∆ ABC AND AC

➨∠ AEB = ∠ AFC ( 90° )

➨ ∠ BAE = ∠ CAF ( Common )

➨ and BE = CF

➨ By ∆ AS

➨ ∆ ABE = ∆ ACF

➨ AB = AC

★ FINAL ANSWER :

• BE = CF ( By C P C T )

∴ Hence, ∆ ABC is isosceles triangle.