Q6 of 31 Page 118

In Fig. 7.21, AC = AE, AB = AD and∠ BAD = ∠ EAC. Show that BC = DE.

It is given in the question that:

AC = AE


AB = AD and,


∠BAD = ∠EAC


To show: BC = DE


Proof:


∠BAD = ∠EAC (Adding ∠DAC both sides)


∠BAD + ∠DAC = ∠EAC + ∠DAC


∠BAC = ∠EAD


In


AC = AE (Given)


∠BAC = ∠EAD


AB = AD (Given)


Therefore,


By SAS congruence,



BC = DE (By c.p.c.t)


More from this chapter

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4

l and m are two parallel lines intersected by another pair of parallel lines p and q(see Fig. 7.19). Show that Δ ABC ≅Δ CDA.


 5

Line l is the bisector of an angle ∠ A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠ A (see Fig. 7.20). Show that:

(i) Δ APB ≅Δ AQB


(ii) BP = BQ or B is equidistant from the arms of ∠ A.


 7

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 Fig. 7.22). Show that

(i) Δ DAP ≅Δ EBP


(ii) AD = BE


 8

In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B(see Fig. 7.23). Show that:

(i) Δ AMC ≅Δ BMD


(ii) ∠DBC is a right angle.


(iii) Δ DBC ≅Δ ACB


(iv) CM =AB