Q3 of 39 Page 158

In Δ ABC, AB > AC; then the bisector of ∠BAC intersects BC at D. Let’s prove that BD < CD.


Given, AD is the bisector of ∠A of triangle ABC .


Hence, ∠DAB = ∠DAC


Now, it is given that, AB > AC


Therefore,


∠BDA > ∠ ADC


Hence, BD < CD


Proved.


More from this chapter

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1

In figure ∠QPR > ∠PQR. Let’s write the relation between PR and QR


 2

In ΔABC, AC > AB. D is any point on the side AC such that ∠ADB = ∠ABD. Let’s prove that ∠ABC > ∠ACB.

 4

In ΔABC, AD is perpendicular to BC and AC > AB. Prove that

(i) ∠CAD > ∠BAD


(ii) DC > BD


 5

In a quadrilateral the greatest and the smallest sides are opposite to each other. Let’s prove that the adjacent angle to the greatest side is smaller than its opposite side.