Areas of Parallelograms and Triangles
Given: AD and BC are equal perpendiculars to a line segment AB.
To Prove: CD bisects AB.
Proof: In ∆O AD and ∆OBC
AD = BC | Given
∠OAD = ∠OBC | Each = 90°
∠AOD = ∠BOC
| Vertically Opposite Angles
∴ ∠OAD ≅ ∆OBC | AAS Rule
∴ OA = OB | C.P.C.T.
∴ CD bisects AB.
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(i) ∆DAP ≅ ∆EBP
(ii) AD = BE.
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.
In figure given below, AD is the median of ∆ABC.
BE ⊥ AD, CF ⊥ AD. Prove that BE = CF.
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