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Midpoint Theorem on Right-angled Triangle, Proof, Statement

Midpoint Theorem on Right-angled Triangle, Proof, Statement

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$Midpoint Theorem on Right-angled Triangle, Proof, Statement$

Here we will prove that in a right-angled triangle the median drawn to the hypotenuse is half the hypotenuse in length. Solution: In ∆PQR, ∠Q = 90°. QD is the median drawn to hypotenuse PR

Midpoit Theorem.pdf

SOLVED: Given: N is the midpoint of MP, QN = RN Prove: AMSP is isosceles Statements (1) QN = RN (Given) (2) N is the midpoint of MP (Given) (3) LMQN and

Converse of Midpoint Theorem Proof of Converse of Midpoint Theorem

PROOF Complete the coordinate proof for the statement. In an isosceles right triangle, the [coordinate geometry]

Answered: A. Complete the two-column proof of the…

Midpoint Theorem on Right-angled Triangle, Proof, Statement

SOLVED: Statements: 1. Given LMER is a right triangle with ZMER as the right angle and MR as the hypotenuse. 2. EY is an altitude to the hypotenuse of AMER. Prove: AMER

$\\Solution:\\\\ \text{Given: In right angled }\Delta ABC, \angle B=90^0\text{ and D is the midpoint of AC}\\\\ RTP:BD=\frac12AC\\\\ Construction: \text{Draw a line DE such that}DE\parallel AB\\\\ Proof:\\ \text{In triangle ABC D is the mid point of AC and }AB \parallel DE\\\\ \Rightarrow \text{E is the mid point}\ \ \ \ \ (\because \text{Converse of midpoint theorem})\\\\ \therefore BE=EC\\\\ From\ \Delta BDE\ and\ \Delta CDE\\\\ \angle DEB=\angle DEC=90^0\\\\ \text{DE is common side}\\\\ \therefore \Delta BDE\cong \Delta CDE\ \ \ \ \ \ \ (\because \text{By SAS congruence rule})\\\\ So,\ BD=DC\ \rightarrow (1)\ \ \ \ \ \ (CPCT)\\\\ DC=\frac12AC\ \rightarrow (2)\ \ \ \ \ \ \ (\because \text{D is the midpoint of AC})\\\\ \text{From (1) and (2)}\\\\ \therefore BD=\frac12AC\\\\ \text{Hence proved}$

Line segment joining the midpoint of the hypotenuse Areas of Parallelograms and Triangles-Maths-Class-9

How to prove the midpoint of the hypotenuse of a right angled triangle is equidistant from