Question

In Triangle ABC, angle B = 90°. M and N are mid-points
of AB and BC respectively. Prove that
(i) CM2 + AN2 = 5MN2
(ii) AN2 + CM? = AC2 + MN2​

Answer 1

(i) correct your problem , and than see solution in image .

Answer 2

Answer with Step-by-step explanation:

In triangle ABC , angle B=90 degrees

AM=BM

BN=NC

AB=AM+MB=BM+BM=2BM

BC=$BN+NC=BN+BN=2BN$

$AC^2=AB^2+BC^2$

$MN^2=BN^2+BM^2$

$AN^2=AB^2+BN^2$

$MC^2=BM^2+BC^2$

Using Pythagoras theorem

$(Hypotenuse)^2=(Base)^2+(perpendicular\;side)^2$

1.$CM^2+AN^2=AB^2+BN^2+BM^2+BC^2$

$CM^2+AN^2=(2BM)^2+BN^2+(2BN)^2$

$CM^2+AN^2=4BM^2+BN^2+4BN^2+BM^2$

$CM^2+AN^2=5BM^2+5BN^2=5(BM^2+BN^2)=5MN^2$

Hence, proved.

2.$AN^2+CM^2=(AB^2+BC^2)+(BN^2+BM^2)$

$AN^2+CM^2=AC^2+MN^2$

Hence, proved.

#Learns more:

https://brainly.in/question/14471954:answered by manisha patel