In Figure 6, two circles touch each other at A. A common tangent touches
them at B and C and another common tangent at A meets the previous
common tangent at P. Prove that BAC = 90°.
Answers
Answer:
Step-by-step explanation:
AP=PC(tangents from external point are equal in length)
AP=BP (tangents from external point are equal in length)
THEREFORE, PC=BP
THEREFORE, <PAC= <PAB ( <s opp to equal sides are equal)
SINCE, BP=AP=AC
<PBA=<PAB(=<PAC)=<PCA
THEREFORE, <PBA=<PCA
CONSIDER <PBA=<PCA=X
IN TRIANGLE BAC,
<BAC+<CBA+<BCA =180(ANGLE SUM PROPERTY).....(I)
SINCE TANGENTS ARE PERPENDICULAR TO RADIUS,
<CBA=90-X .......(II)
SIMILARLY, <BCA=90-X .......(III)
SUBSTITUTING (II),(III) IN (I)
<BAC+90-X+90-X=180
<BAC=180-180+2X
<BAC=2X .......(IV)
THEREFORE, <BAC+<CBA+<BCA=X+X+2X=180
4X=180
X=45
SUBSTITUTING X IN (IV)
<BAC=2(45)
<BAC= 90°
Answer:
GIVEN: TWO CIRCLES AND COMMON TANGENTS BC AND PA
.
TO PROVE: <BAC=90°
Step-by-step explanation:
AP=PC(tangents from external point are equal in length)
AP=BP (tangents from external point are equal in length)
THEREFORE, PC=BP
THEREFORE, <PAC= <PAB ( <s opp to equal sides are equal)
SINCE, BP=AP=AC
<PBA=<PAB(=<PAC)=<PCA
THEREFORE, <PBA=<PCA
CONSIDER <PBA=<PCA=X
IN TRIANGLE BAC,
<BAC+<CBA+<BCA =180(ANGLE SUM PROPERTY).....(I)
SINCE TANGENTS ARE PERPENDICULAR TO RADIUS
<CBA=90-X .......(II)
SIMILARLY, <BCA=90-X .......(III)
SUBSTITUTING (II),(III) IN (I)
<BAC+90-X+90-X=180
<BAC=180-180+2X
<BAC=2X .......(IV)
THEREFORE, <BAC+<CBA+<BCA=X+X+2X=180
4X=180
X=45
SUBSTITUTING X IN (IV)
<BAC=2(45)
<BAC= 90°