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Question 4 on page 23 of mathematics book in grade 2 of junior high school
Because ad bisectors BAC,
AB is parallel to PE, AC is parallel to PF,
So PD bisector PEF,
So the distance from D to PE is equal to the distance from D to PF
People's education press, page 23, question 4, a little bit do not understand ~
Because ad bisectors BAC,
AB is parallel to PE, AC is parallel to PF,
So PD bisector PEF,
So the distance from D to PE is equal to the distance from D to PF
The answer to question 9 on page 27 of congruent triangle chapter in mathematics textbook for grade two of people's education (30marks)
Yes, this is our homework today
Because the angle ABC = 90 degree, the angle BCE + the angle DCA = 90 degree
Be perpendicular to CE ad perpendicular to CE
And because angle BCE + angle CBE = 90 degree
So angle CBE = angle ACD
Because the angle BEC = angle ADC = 90 degrees
AC=BC
So triangle BEC is equal to triangle ADC
So ad = CE, because CE = CD = De, CD = be
So be = ad-de = 2.5cm-1.7cm = 0.8cm
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