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

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.

Answers (1)

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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