Q11 of 49 Page 13

In ΔABC, ray BD bisects ∠ABC and ray CE bisects ∠ACB. If seg AB ≅ seg AC then prove that ED || BC.

PROOF:



Theorem: The bisector of an angle of a triangle divides the side opposite to the angle in the ratio of the remaining sides.


⇒ ……(1)


and …..(2) (∵ , BD and CE are angle bisectors of ∠B and ∠C respectively.)


Now, ∵ seg AB ≅ seg AC


⇒ AB = AC


⇒


⇒ R.H.S of (1) & (2) are equal.


⇒ L.H.S of (1) & (2) will be equal.


∴ Equating L.H.S of (1) &(2), we get-


⇒


⇒ ED||BC (By converse basic proportionality theorem)


More from this chapter

All 49 →
9

In Δ ABC, seg BD bisects ∠ ABC. If AB = x, BC = x + 5, AD = x – 2, DC = x + 2, then find the value of x.

 10

In the figure 1.44, X is any point in the interior of triangle. Point X is joined to vertices of triangle. Seg PQ || seg DE, seg QR || seg EF. Fill in the blanks to prove that, seg PR || seg DF.


Proof : IN



(Basic proportionality theorem)


In



(converse of basic proportionality theorem)

 1

In figure 1.55, ∠ABC = 75°, ∠EDC = 75° state which two triangles are similar and by which test? Also write the similarity of these two triangles by a proper one to one correspondence.

 2

Are the triangles in figure 1.56 similar? If yes, by which test?