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ABC is an isosceles triangle in which AC = BC. AD and BE are respectively two altitudes to sides BC and AC. Prove that AE = BD.

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Given: \triangleABC is isosceles triangle AB = AC

AD and BE are altitudes at BC and AC

To Prove: AE = BD

Proof: In \triangleABD & \triangleACD

AD = AD                                             (common)

\angleADB = \angleADC                                (AD is altitude at BC)

AB = AC                                             (Given)

\therefore \triangleABD \cong \triangleACD                  (by SAS congruence)                        

\angleBAD = \angleDAC                                (by CPCT)

\angleBAD = \angleDAE                      (ÐDAC = ÐDAE)                   (i)

Now, in \triangle ABD and \triangle ABE

AB = BA                                             (common)

\angleADB = \angleAEB = 90°            (AD and BE are altitudes)

\angleBAD = \angleDAE                      (From i)

\therefore \triangle ABD \cong \triangle ABE                   (by AAS congruency)

\therefore AE = BD                                          (by CPCT)

Hence proved.

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