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andDBDC,i.e.,In the given figure, AB AC<DBC = <DCB. Prove that LABD = ACD and <BA)-CAD, i.e., AD bisects LBAC of a AABC. ICBSE 2010] |
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Answer» inΔABCAB=AC∴∠ABC=∠ACB(angles opposite to equal sides of a triangle are equal)......1inΔDBC,DB=DC,∴∠DBC=∠DCB(angles opposite to equal sides of a triangle are equal)......2subtract 2 from 1∠ABC-∠DBC=∠ACB-∠DCB(equals subtracted from equals gives equal)=∠ABD=∠ACDdivide both the sides by∠ACD⇒∠ABD/∠ACD=1∴∠ABD:∠ACD=1:1 |
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