As shown in the figure, in the triangle ABC, D is the midpoint of BC, ad is perpendicular to BC at point D, De is perpendicular to ab at point E, de = 5cm, calculate the distance from point d to AC
Solution ∵ BD = DC ∠ ADB = ∠ ADC = 90 ° with AD = ad
∴△ABD≌△ADC ∴∠BAD=∠CAD
∵ de = 5 ∵ the distance from point d to AC is equal to 5 (the distance from the point on the bisector to both sides of the angle is equal)
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