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Areas of Parallelograms and Triangles
Question
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In figure given below, AD is the median of ∆ABC.
BE ⊥ AD, CF ⊥ AD. Prove that BE = CF.

Answer

Given: AD is the median of ∆ABC. BE ⊥ AD, CF ⊥ AD.
To Prove: BE = CF
Proof: In ∆DEB and ∆DFC,
DB = DC
| ∵ AD is the median of ∆ABC
∠DEB = ∠DFC | Each = 90°
∠BDE = ∠CDF
| Vertically opposite angles
∴ ∆DEB ≅ ∆DFC
| AAS congruence rule
∴ BE = CF    | CPCT

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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.

In figure, diagonal AC of a quadrilateral ABCD bisects the angles A and C. Prove that AB = AD and CB = CD.

AB is a line-segment. AX and BY are two equal line-segments drawn on opposite sides of line AB such that AX || BY. If AB and XY intersect each other at P. Prove that:
(i)    ∆APX ≅ ∆BPY
(ii)    AB and XY bisect each other at P.

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In the given figure, if AB = FE, BC = ED, AB ⊥ BD and FE ⊥ EC, then prove that AD = FC.

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