conditions
TYPE-1: Questions based on SAS
1. Prove that AABC is isosceles if any one
of the following holds:
(1) Bisector of ZBAC is perpendicular to
B
BC
(ii) Altitude AD bisects ZBAC
OJAI
MAltitude BE = Altitude CF
AD
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Answers
Answer:
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Step-by-step explanation:
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Answer:
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Step-by-step explanation:
A) __
Altitude AD bisects ∠BAC
Altitude AD bisects ∠BAC
Altitude AD bisects ∠BAC In △BAD and △CAD
Altitude AD bisects ∠BAC In △BAD and △CADAD=AD (Common Side)
Altitude AD bisects ∠BAC In △BAD and △CADAD=AD (Common Side)∠ADB=∠ADC=90 o
(AD is altitude)
(AD is altitude)∠BAD=∠CAD (AD bisects ∠BAC)
(AD is altitude)∠BAD=∠CAD (AD bisects ∠BAC) ∴△BAD≅△CAD(A.S.A)
(AD is altitude)∠BAD=∠CAD (AD bisects ∠BAC) ∴△BAD≅△CAD(A.S.A)⇒AB=AC(C.P.C.T.C.)
(AD is altitude)∠BAD=∠CAD (AD bisects ∠BAC) ∴△BAD≅△CAD(A.S.A)⇒AB=AC(C.P.C.T.C.)Thus, △BAC is an isosceles triangle.
(AD is altitude)∠BAD=∠CAD (AD bisects ∠BAC) ∴△BAD≅△CAD(A.S.A)⇒AB=AC(C.P.C.T.C.)Thus, △BAC is an isosceles triangle.B) __
Median AD is perpendicular to BC
Median AD is perpendicular to BC
Median AD is perpendicular to BC
Median AD is perpendicular to BC In △BAD and △CAD
Median AD is perpendicular to BC In △BAD and △CADAD=AD (Common Side)
Median AD is perpendicular to BC In △BAD and △CADAD=AD (Common Side)∠ADB=∠ADC=90 o
(AD is perpendicular to BC)
(AD is perpendicular to BC)BD=CD (AD is Median)
(AD is perpendicular to BC)BD=CD (AD is Median) ∴△BAD≅△CAD(S.A.S.)
(AD is perpendicular to BC)BD=CD (AD is Median) ∴△BAD≅△CAD(S.A.S.)⇒AB=AC(C.P.C.T.C.)
(AD is perpendicular to BC)BD=CD (AD is Median) ∴△BAD≅△CAD(S.A.S.)⇒AB=AC(C.P.C.T.C.)Thus, △BAC is an isosceles triangle.
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