Can anyone ans both the ques plz....
Answers
Answer:
Step-by-step explanation:
(1). Given:P,Q,R are the mid points of sides BC,CA and AB of a triangle ABC. PR and BQ meet at X. CR and PQ meet at Y.
To Prove: XY = 1/4 BC
Proof: In Δ ABC, Q and R are the mid-points of sides AC and AB respectively.
QR = 1/2 BC........(1).
⇒QR = BP [P is the mid-point of BC].
Using Mid-point Theorem, it can also be said that QR || BC.
⇒ QR || BP
In quadrilateral BPQR, BP || QR and BP = QR.
Thus, BPQR is a parallelogram
Now, the diagonals BQ and PR of the parallelogram BPQR bisect each other at X.
Thus, X is the mid-point of PR
Similarly, it can be formed that Y is the mid-point of PQ
In ΔPQR, X and Y are the mid-points of sides PR and PQ respectively.
=> XY = 1/2QR
=> XY = 1/2*1/2 BC
=> XY = 1/4 BC
Hence, proved.
(2). Given: ∠AEB=x, ∠APB=y and ∠AOB=z.
To prove: ∠x+∠y=∠z.
As shown in figure,
Let ∠ACB=c and ∠ADB=d.
∴ ∠PCE=(180°−c) and ∠PDE=(180°−d)
So in quadrilateral, EPCD
∠CED+∠EDC+∠DPC+∠PCE=360°
(As we know, sum of all angle of quadrilateral is 360°)
x+ (180°−d)+y+(180°−c)=360°
x+y=360°−360°+c+d
x+y=c+d (1).
We, know that, ∠AOB=2∠ACB (By the theorem, the angle formed at the center of the circle by the linesoriginating from two points on the circle's circumference is double the angle formedon the circumference of the circle by lines originating from the same points.)
∠AOB=2∠ACB⇒ z=2c (2). ∠AOB=2∠ADB⇒ z=2d (3).
By adding (2) and (3),
2c+2d=2z. (4).
⇒z=c+d From equation (1) and (4),
Hence proved, ∠x+∠y=∠z