मराठी

ΔABC and ΔDBC lie on the same side of BC, as shown in the figure. From a point P on BC, PQ || AB and PR || BD are drawn, meeting AC at Q and CD at R respectively. Prove that QR || AD.

Advertisements
Advertisements

प्रश्न

ΔABC and ΔDBC lie on the same side of BC, as shown in the figure. From a point P on BC, PQ || AB and PR || BD are drawn, meeting AC at Q and CD at R respectively. Prove that QR || AD. 

 

सिद्धांत
Advertisements

उत्तर

In Δ CAB, PQ || AB.
Applying Thales' theorem, we get: 

`(CP)/(PB)=(CQ)/(QA)`                   ...............(1) 

Similarly, applying Thales theorem in BDC , Where PR||DM we get:  

`(CP)/(PB)=(CR)/(RD)`                  ..................(2) 

Hence, from (1) and (2), we have : 

`(CQ)/(QA)=(CR)/(RD)` 

Applying the converse of Thales’ theorem, we conclude that QR ‖ AD in Δ ADC. This completes the proof.  

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 7: Triangles - EXERCISE 7A [पृष्ठ ३७३]

APPEARS IN

आर. एस. अग्रवाल Mathematics [English] Class 10
पाठ 7 Triangles
EXERCISE 7A | Q 9. | पृष्ठ ३७३
Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×