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Areas Of Parallelograms And Triangles

Question

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In ∆ABC, ∠B = 35°, ∠C = 65° and the bisector of ∠BAC meets BC in X. Then, the relation between BX and AX is

BX = AX
BX < AX
BX > AX
none of these

Solution

C.

BX > AX

Some More Questions From Areas of Parallelograms and Triangles Chapter

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.

In figure given below, AD is the median of ∆ABC.
BE ⊥ AD, CF ⊥ AD. Prove that BE = CF.

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

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