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Areas of Parallelograms and Triangles
Question
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Prove that the angles opposite to equal sides of a triangle are equal. Is the converse true? 

Answer

Given: A triangle ABC in which AB = AC
To Prove: ∠ABC = ∠ACB

Construction: Draw the bisector AD of A so as to intersect BC at D.
Proof: In ∆ADB and ∆ADC,
AD = AD    | Common
AB = AC    | Given
∠BAD = ∠CAD
| By Construction
∴ ∆ADB ≅ ∆ADC
| SAS congruence rule
∴ ∠ABD = ∠ACD    | CPCT
⇒ ∠ABC = ∠ACB
Yes, the converse is true.

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Some More Questions From Areas of Parallelograms and Triangles Chapter

In figure, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.

AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA = ∠DPB (see figure). Show that.

(i) ∆DAP ≅ ∆EBP
(ii) AD = BE.

 

In figure, PS = QR and ∠SPQ = ∠RQP. Prove that PR = QS and ∠QPR = ∠PQS.

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.

In figure, ∠QPR = ∠PQR and M and N are respectively points on sides QR and PR of ∆PQR, such that QM = PN. Prove that OP = OQ, where O is the point of intersecting of PM and QN.

In figure, ∠B = ∠.E, BD = CE and ∠1 = ∠2. Show ∆ABC ≅ ∆AED.