Areas of Parallelograms and Triangles
Given: D and E are points on the base BC of a ∆ABC such that AD = AE and ∠BAD = ∠CAE.
To Prove: AB = AC
Proof: In ∆ADE,
∵ AD = AE | Given
∴ ∠ADE = ∠AED ...(1)
| Angles opposite to equal sides of a triangle are equal
In ∆ABD,
Ext. ∠ADE = ∠BAD + ∠ABD ...(2)
| An exterior angle of a triangle is equal to the sum of its two interior opposite angles
In ∆AEC,
Ext. ∠AED = ∠CAE + ∠ACE .. .(3)
| An exterior angle of a triangle is equal to the sum of its two interior opposite angles
From (1), (2) and (3),
∠BAD + ∠ABD = ∠CAE + ∠ACE
⇒ ∠ABD = ∠ACE
| ∵ ∠BAD = ∠CAE (Given)
⇒ ∠ABC = ∠ACB
∴ AB = AC
| Sides opposite to equal angles of a triangle are equal
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Line I is the bisector of an angle ∠A and B is any point on I. BP and BQ are perpendiculars from B to the arms of ∠A (see figure). Show that:
(i) ∆APB ≅ ∆AQB
(ii) BP = BQ or B is equidistant from the arms of ∠A.
(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.
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