Q5 of 23 Page 187

ABC is an isosceles triangle in which AB = AC. AD bisects exterior angle QAC and CD || BA as shown in the figure. Show that

(i) ∠DAC = ∠BCA


(ii) ABCD is a parallelogram



Given:- AB=AC and ABC is isosceles triangle


∠ QAD=∠ DAC


CD||BA


Formula used:- sum of angles of triangle is 180°


Solution:-


As Δ ABC is a isosceles triangle


∠ ABC=∠ ACB


∠ ABC+∠ ACB+∠ CAB=180°


2∠ ACB+∠ CAB=180°


*As BAQ is a straight line


∠ CAB+∠ DAC+∠ QAD=180°


*∠ QAD=∠ DAC [Given]


∠ CAB=180° - 2∠ DAC


Putting value of ∠ CAB in above equation 1


2∠ ACB+180° - 2∠ DAC=180°


2∠ ACB =2∠ DAC


∠ ACB =∠ DAC


If;


There is ∠ ACB =∠ DAC and AC is the transverse


∴ these are equal by alternate angles


And AD||BC


In ABCD


If;


AD||BC & CD||BA


If both pair of sides of quadrilateral are parallel


Then the quadrilateral is parallelogram


Conclusion:-


ABCD is a parallelogram


And ∠DAC = ∠BCA


More from this chapter

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3

In the adjacent figure ABCD is a parallelogram and E is the midpoint of the side BC. If DE and AB are produced to meet at F, show that AF = 2AB.


 4

In the adjacent figure ABCD is a parallelogram P, Q are the midpoints of sides AB and DC respectively. Show that PBCQ is also a parallelogram.


 6

ABCD is a parallelogram AP and CQ are perpendiculars drawn from vertices A and C on diagonal BD (see figure) show that

(i) Δ APB ≅ Δ CQD


(ii) AP = CQ



 7

In and DEF, AB || DE, AB=DE; BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see figure). Show that

(i) ABED is a parallelogram


(ii) BCFE is a parallelogram


(iii) AC = DF


(iv) ∆ABC ≅ ∆DEF