Prove that in a right triangle, the square of the hypotenuse is equal to sum of squares of other two
sides. Using the above result, prove that, in rhombus ABCD, 4 AB2= AC2+ BD2
Answers
Answer:
rhombus one is left..... wait for 5 mins
Answer:
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Step-by-step explanation:
Given: In ΔABC, m∠ABC=90°
Construction: BD is a perpendicular on side AC
To Prove: (AC)²=(AB)²+(BC)²
Proof:
In △ABC,
m∠ABC=90° (Given)
Since BD is perpendicular to hypotenuse AC (Construction)
Therefore, △ADB∼△ABC∼△BDC (Similarity of right-angled triangle)
△ABC∼△ADB
(AB/AC)=(AD/AB) (congruent sides of similar triangles)
AB2=AD×AC (1)
△BDC∼△ABC
CD/BC=BC/AC (congruent sides of similar triangles)
BC2=CD×AC (2)
Adding the equations (1) and (2),
AB2+BC2=AD×AC+CD×AC
AB2+BC2=AC(AD+CD)
Since, AD + CD = AC
Therefore, AC2=AB2+BC2
Hence Proved.
Given: A rhombus ABCD
To Prove: 4AB2 = AC2 + BD2
Proof: The diagonals of a rhombus bisect each other at right angles.
OA=1/2 AC OB=1/2BD
Angle AOB = 90
In triangle AOB,
AB2= OB2+OA2
AB2= BD/2^2+AC/2^2
AB2= AC2 + BD2
Therefore,
4AB2+ AC2 + BD2
Hence proved...