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Areas of Parallelograms and Triangles
Question
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In figure, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.

Answer

Given: In figure, AC = AE, AB = AD and ∠BAD = ∠EAC.
To Prove: BC = DE.
Proof: In ∆ABC and ∆ADE,
AB = AD    | Given
AC = AE    | Given
∠BAD = ∠EAC    | Given
⇒ ∠BAD + ∠DAC = ∠DAC + ∠EAC
| Adding ∠DAC to both sides
⇒    ∠BAC = ∠DAE
∴ ∆ABC ≅ ∆ADE | SAS Rule
∴ BC = DE.    | C.P.C.T.

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Some More Questions From Areas of Parallelograms and Triangles Chapter

In figure, AP and BQ are perpendiculars to the line-segment AB and AP = BQ. Prove that O is the midpoint of line segments AB and PQ.

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