Answer:
Step-by-step explanation:
Answer:
TRUE
Step-by-step explanation:
Given that line segment JK and LM are parallel. From picture we see that LK is transversal line.
We know that corresponding angles formed by transversal line are congruent.
Hence ∠JKL = ∠ MLK ...(i)
Now consider triangles JKL and MLK
JK = LM {Given}
∠JKL = ∠ MLK { Using (i) }
KL = KL {common sides}
Hence by SAS property of congruency of triangles, ΔJKL and ΔMLK are congurent.
Hence given statement is TRUE.
Answer:
CR = 17 ; PR = 30
Step-by-step explanation:
From the problem hypothesis we know that CD⊥PR or CQ⊥PR(perpendicular) and ∡PCQ ≡∡RCQ (bisector)
so ∡PCQ≡∡RCQ
[CQ]≡[CQ] (common side)
CQ⊥PR
--------------------------------------------
⇒(cathetus - angle case of congruence)⇒ ΔPQC≡ΔRQC so [PQ]≡[QR] (=15) // PR = PQ+QR ⇒ PR=30
from this congruence ⇒ΔPRC = isosceles so PC=CR=17
Hint: in any triangle is bisector of an angle it is perpendicular on the opposite side then triangle is isosceles.
The angle is 42 degrees and its complement is 48 degrees
