Answer:
Step-by-step explanation:
6). Since, m∠JMN = 90° [Given]
And m∠JMK + m∠NMK = m∠JMN
m∠JMK = m∠JMN - m∠NMK
= 90° - 28°
= 62°
From the other figure,
m∠STR + m∠STP = 180° [Linear pair of angles]
m∠STR = 180° - m∠STP
= 180° - 118°
= 62°
Therefore, mJMK = m∠STR = 62°
And m∠JMK ≅ m∠STR
8). Since, BF is the angle bisector of ∠AFC,
∠AFB ≅ ∠BFC [By angle bisector theorem]
∠CFD ≅ ∠BFC [Given]
Therefore, ∠AFB ≅ ∠CFD [Transitive property of equality]
9). AG is the bisector of CD,
Therefore, CE ≅ DE ------- (1)
IJ is the bisector of CE,
Therefore, CK ≅ KE ---------(2)
BH is the bisector of ED,
Therefore, EF ≅ FD -------(3)
Since, CE ≅ DE [Given as (1)]
(CK + KE) ≅ (EF + FD)
2(KE) ≅ 2(EF) [Given in properties (2) and (3)]
KE ≅ EF
